Question

Stie cineva tema, va rog :(

Answer 1

n=m/M=3,12/104 = 0,03 moli s.o

1dm3=1L

1g=1000mg

m/M=V/Vm=>m=M*V/Vm=44*3,36/22,4=6,6g CO2

m H2O=3,24g

44g CO2......................12g C
6,6g CO2......................x=1,8g C

18g H2O.....................2g H2
3,24g H2O.................x=0,36g H2

m O =3,12-1,8-0,36=0,96g

n=m/M=1,8/12=0,15 moli C

n=m/M=0,36/2=0,18 moli H2

n=m/M=0,96/16=0,06 moli O

0,03 moli s.o................0,15 moli C..........0,18 moli H2.......0,06 moli O
1 mol...........................x=5 atomi C..........y=12 atomi H2......z=2 atomi O

F.M => C5H12O2

sau

%C=300a/11s=300*6,6/11*3,12=57,6923

%H=100b/9s=100*3,24/9*3,12=11,538

%O=100-(%C+%H)=30,76923

nC = %C*Ms.o/AC*100=57,6923*104/1200= 5 atomi C

nH=%H*Ms.o/AH*100=11,538*104/100=12 atomi H

nO=%O*Ms.o/AO*100=30,76923*104/1600=2 atomi O

F.M => C5H12O2